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Popular >

extreme f(x)=2x^2+x+2

Solution

extreme points

f (x) = 2 x^{2} + x + 2

Solution

Minimum (- \frac{1}{4}, \frac{15}{8})

Solution steps

f^{'} (x) = 4 x + 1

Find intervals:

Decreasing

: - \infty < x < - \frac{1}{4},

Increasing

: - \frac{1}{4} < x < \infty

Plug

x = - \frac{1}{4}

into

2 x^{2} + x + 2 : \frac{15}{8}

Minimum (- \frac{1}{4}, \frac{15}{8})

Graph

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Frequently Asked Questions (FAQ)

What is the extreme f(x)=2x^2+x+2 ?
The extreme f(x)=2x^2+x+2 is Minimum(-1/4 , 15/8)